Find
\[\sum_{n = 1}^\infty \frac{2^n}{1 + 2^n + 2^{n + 1} + 2^{2n + 1}}.\]
Solution: First, we can factor the denominator:
\[1 + 2^n + 2^{n + 1} + 2^{2n + 1} = (1 + 2^n) + 2^{n + 1} (1 + 2^n) = (1 + 2^n)(1 + 2^{n + 1}).\]Then we can write the numerator $2^n$ as $(1 + 2^{n + 1}) - (1 + 2^n) = 2^n,$ so
\[\frac{2^n}{1 + 2^n + 2^{n + 1} + 2^{2n + 1}} = \frac{(1 + 2^{n + 1}) - (1 + 2^n)}{(1 + 2^n)(1 + 2^{n + 1})} = \frac{1}{1 + 2^n} - \frac{1}{1 + 2^{n + 1}}.\]Hence,
\begin{align*}
\sum_{n = 1}^\infty \frac{2^n}{1 + 2^n + 2^{n + 1} + 2^{2n + 1}} &= \left( \frac{1}{1 + 2} - \frac{1}{1 + 2^2} \right) + \left( \frac{1}{1 + 2^2} - \frac{1}{1 + 2^3} \right) + \left( \frac{1}{1 + 2^3} - \frac{1}{1 + 2^4} \right) + \dotsb \\
&= \boxed{\frac{1}{3}}.
\end{align*}